Question

A toy rocket is shot vertically upward from the ground. Its distance in feet from the ground in t seconds is given by s(t) = -16t^2 + 142t. At what time or times will the ball be 140 ft from the ground? round your answer to the nearest tenth, if necessary.

Answer 1

Answer:

The ball will be 140 ft from the ground when t = 1.1s and when t = 7.7s

Step-by-step explanation:

To solve this question, we need to understand how to solve a quadratic equation.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

The position of the pall is given by the following equation.

$s(t) = -16t^{2} + 142t$

At what time or times will the ball be 140 ft from the ground?

This is t when $s(t) = 140$

So

$140 = -16t^{2} + 142t$

$16t^{2} - 142t + 140 = 0$

So

$a = 16, b = -142, c = 140$

$\bigtriangleup = b^{2} - 4ac = (-142)^{2} - 4*16*140 = 11204$

$t_{1} = \frac{-(-142) + \sqrt{11204}}{2*16} = 7.7$

$t_{2} = \frac{-(-142) - \sqrt{11204}}{2*16} = 1.1$

The ball will be 140 ft from the ground when t = 1.1s and when t = 7.7s